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Math Help - race problem

  1. #1
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    race problem

    In a race of 1000 meters, A beats B by 200 meters and A beats C by 500 meters. Assuming that the contestants run at constant speeds, by how many meters does B beat C?

    (A) 300 (B) 375 (C) 450 (D) 500 (E) 625
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  2. #2
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    Quote Originally Posted by sri340 View Post
    In a race of 1000 meters, A beats B by 200 meters and A beats C by 500 meters. Assuming that the contestants run at constant speeds, by how many meters does B beat C?

    (A) 300 (B) 375 (C) 450 (D) 500 (E) 625
    A speed = \frac{1000}{t}

    B speed = \frac{800}{t}

    C speed = \frac{500}{t}

    After t, B has 200 metres to travel, an extra quarter of his distance already travelled in t (not a fifth of 1000, it's a quarter of 800).
    This will take him an extra \frac{t}{4}

    In \frac{t}{4}, C will have travelled a quarter of 500,

    which is 125 metres.

    Therefore C still has 1000-(500-125) metres to travel, when B crosses the line.
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  3. #3
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    Hello, sri340!

    Another approach (very primitive) . . .


    In a 1000-m race, A beats B by 200 meters and A beats C by 500 meters.
    Assuming that the contestants run at constant speeds, by how many meters does B beat C?

    . . (A)\;300 \qquad(B)\;375 \qquad (C)\;450 \qquad (D)\;500 \qquad (E)\; 625

    The race ended like this:
    Code:
                              C           B       A 
          + - - - - - - - - - o - - - - - o - - - o
          0                  500         800    1000

    C ran 500 m in the time that B ran 800 m.
    C's speed is \tfrac{500}{800} = \tfrac{5}{8} of B's speed.

    When B ran his last 200 m, C ran only: . \tfrac{5}{8}(200) \,=\,125 m.

    As B crosses the finish line, C is at the 625-m mark.


    Therefore, B beats C be 375 m . . . answer (B).

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